diketahui 2log3=a dan 3log5=b .nilai 6log40 adalah

6log40 = (2log40)/(2log6) =
(2log(2³.5)) / (2log(2.3)) =
(2log2³ + 2log5) / (2log2 + 2log3) =
( 3 + 2log3 × 3log5 ) / ( 1 + 2log3 ) =
( 3 + a × b ) / ( 1 + a ) =
3ab/(a+1)
Jadi, hasilnya 3ab/(a+1)

6 log 40 =
3 log 40/ 3 log 6
3 log 2×2×2×5 / 3 log 2×3
3/a + b per 1/a + 1
3+ab/a per 1+a/a
3+ab/a × a / 1+a
3+ab per 1+a